Question

In an ideal solenoid there are $330$ turns with a length of $23\text{cm}$ and a radius of cross- section $1.3\text{cm}$. It carries a current of $2.3\text{A}$. The current in the $330$ turns solenoid increases steadily to $5.00\text{A}$ in $0.9\text{s}$. Calculate the magnetic field of the $330$ turns solenoid after $0.9\text{s}$.

• A. $0.9\times {10}^{-3}\text{T}$
• B. $0.09\times {10}^{-4}\text{T}$
• C. $9\times {10}^{-3}\text{T}$
• D. $0.09\times {10}^{-3}\text{T}$

Answer 1

The correct option is C $9\times {10}^{-3}\text{T}$

Given:

$r=1.3\text{cm}=1.3\times {10}^{-2}\text{m};N=330;l=23\text{cm}=0.23\text{m}$

$\therefore n=\frac{330}{0.23}\text{turns/m}$

As given in question, after $0.9\text{s}$, current in the solenoid increases upto $5\text{A}$.

Using formula $B={\mu }_{0}nI$

$B=4\pi \times {10}^{-7}\times \frac{330}{0.23}\times 5$

$\therefore B=9\times {10}^{-3}\text{T}$

Hence, option $\left(C\right)$ is the correct options.

Why this Question? Note: The magnetic field due to an ideal solenoid is given by, $B={\mu }_{0}ni$